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3 years ago

6

What are examples of government action? Please HELP ME!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

1 answer:

alexira [117]3 years ago

6 0

The answer is <span>Encourage private ownership Profit motives and promote competition and monopoly regulate companies sell food and regulate pollution</span>

You might be interested in

Which is not a simple harmonic motion (S.H.M.) (a) Simple Pendulum (b) Projectile motion (c) None (d) Spring motion

Scrat [10]

Answer:

b) Projectile MOTION

Explanation:

SHM is periodic motion or to and fro motion of a particle about its mean position in a straight line

In this type of motion particle must be in straight line motion

So here we can say

a) Simple Pendulum : it is a straight line to and fro motion about mean position so it is a SHM

b) Projectile motion : it is a parabolic path in which object do not move to and fro about its mean position So it is not SHM

d) Spring Motion : it is a straight line to and fro motion so it is also a SHM

So correct answer will be

b) Projectile MOTION

7 0

3 years ago

A cruise ship sails due south at 2.00 m/s while a coast guard patrol boat heads 19.0° north of east at 5.60 m/s. What are the x-

Lilit [14]

Answer:

The x-component and y-component of the velocity of the cruise ship relative to the patrol boat is -5.29 m/s and 0.18 m/s.

Explanation:

Given that,

Velocity of ship = 2.00 m/s due south

Velocity of boat = 5.60 m/s due north

Angle = 19.0°

We need to calculate the component

The velocity of the ship in term x and y coordinate

$v_{s_{x}}=0$

$v_{s_{y}}=2.0\ m/s$

The velocity of the boat in term x and y coordinate

For x component,

$v_{b_{x}}=v_{b}\cos\theta$

Put the value into the formula

$v_{b_{x}}=5.60\cos19$

$v_{b_{x}}=5.29\ m/s$

For y component,

$v_{b_{y}}=v_{b}\sin\theta$

Put the value into the formula

$v_{b_{y}}=5.60\sin19$

$v_{b_{y}}=1.82\ m/s$

We need to calculate the x-component and y-component of the velocity of the cruise ship relative to the patrol boat

For x component,

$v_{sb_{x}}=v_{s_{x}}-v_{b_{x}}$

Put the value into the formula

$v_{sb_{x}=0-5.29$

$v_{sb}_{x}=-5.29\ m/s$

For y component,

$v_{sb_{y}}=v_{s_{y}}-v_{b_{y}}$

Put the value into the formula

$v_{sb_{x}=2.-1.82$

$v_{sb}_{x}=0.18\ m/s$

Hence, The x-component and y-component of the velocity of the cruise ship relative to the patrol boat is -5.29 m/s and 0.18 m/s.

7 0

3 years ago

A roadrunner is running along a straight desert road at a constant velocity of 25 m/s. If a certain coyote wants to capture the

andreyandreev [35.5K]

Answer:

t = 1.42 s and d = 35.5 m

Explanation:

Given that,

Velocity of a roadrunner is 25 m/s

A certain coyote wants to capture the roadrunner using a net dropped from an overpass that is 10 m high.

We need to find the time before the roadrunner is under the overpass and how far away from the overpass is the roadrunner when the coyote drops the net.

$d=ut+\dfrac{1}{2}at^2\\\\\text{Here, u = 0 and a = g}\\\\d=\dfrac{1}{2}gt^2\\\\t=\sqrt{\dfrac{2d}{g}} \\\\t=\sqrt{\dfrac{2\times 10}{9.8}} \\\\t=1.42\ s$

Let d is the distance traveled. So,

d = vt

d = 25 m/s × 1.42 s

d = 35.5 m

5 0

3 years ago

Three identical resistors are connected in parallel. The equivalent resistance increases by 630 when one resistor is removed and

strojnjashka [21]

Answer:

each resistor is 540 Ω

Explanation:

Let's assign the letter R to the resistance of the three resistors involved in this problem. So, to start with, the three resistors are placed in parallel, which results in an equivalent resistance $R_e$ defined by the formula:

$\frac{1}{R_e}=\frac{1}{R} } +\frac{1}{R} } +\frac{1}{R} \\\frac{1}{R_e}=\frac{3}{R} \\R_e=\frac{R}{3}$

Therefore, R/3 is the equivalent resistance of the initial circuit.

In the second circuit, two of the resistors are in parallel, so they are equivalent to:

$\frac{1}{R'_e}=\frac{1}{R} +\frac{1}{R}\\\frac{1}{R'_e}=\frac{2}{R} \\R'_e=\frac{R}{2} \\$

and when this is combined with the third resistor in series, the equivalent resistance ( $R''_e$ ) of this new circuit becomes the addition of the above calculated resistance plus the resistor R (because these are connected in series):

$R''_e=R'_e+R\\R''_e=\frac{R}{2} +R\\R''_e=\frac{3R}{2}$

The problem states that the difference between the equivalent resistances in both circuits is given by:

$R''_e=R_e+630 \,\Omega$

so, we can replace our found values for the equivalent resistors (which are both in terms of R) and solve for R in this last equation:

$\frac{3R}{2} =\frac{R}{3} +630\,\Omega\\\frac{3R}{2} -\frac{R}{3} = 630\,\Omega\\\frac{7R}{6} = 630\,\Omega\\\\R=\frac{6}{7} *630\,\Omega\\R=540\,\Omega$

8 0

3 years ago

Which was least likely to have been a component of Earth’s atmosphere before life began?

Otrada [13]

The answer to this question is a) sulfur

4 0

3 years ago

Read 2 more answers